They talked about this, "In 3-dimensional velocity space, the velocity vectors corresponding to a given speed v live on the surface of a sphere with radius v. The larger v is, the bigger the sphere, and the more possible velocity vectors there are. So the number of possible velocity vectors for a given speed goes like the surface area of a sphere of radius v." Which I completely do not understand.

The velocity space is the 3-dimensional space spanned by vx, vy and vz (given that vx, vy and vz are 3 real numbers, they span a 3-dimensional space ; moreover, velocity is a "good vector", so they even transform correctly - which doesn't matter here).
Now, for a velocity vector (vx,vy,vz), the speed that corresponds to it is:

v = sqrt(vx^2 + vy^2 + vz^2), right ?

So, in that space, all the 3-tupels (vx,vy,vz) that correspond to speed v, are those that satisfy the above equation, which can be re-written:

v^2 = vx^2 + vy^2 + vz^2.

But that's nothing else but the equation for a spherical shell! So all the velocity vectors (vx,vy,vz) that have speed exactly equal to v, lie on a spherical shell, with radius v.
More, consider now all the velocity vectors (vx,vy,vz) such that their speed lies between v and v+dv. Clearly, they have to lie between the shell with radius v and the shell with radius v+dv. But if dv is small, this corresponds to a "layer of paint" with thickness dv onto the shell with radius v. So the volume of paint used for this, is nothing else but the area of the shell, times the thickness of the layer of paint. The area of a spherical shell with radius R is 4 pi R^2, so given that the radius is here v, it becomes 4 pi v^2. And the thickness of the layer is just dv.
So the volume of "velocity space" that corresponds to speeds between v and v+dv is 4 pi v^2 dv.